Page:DeSitterGravitation.djvu/11

we find

${\begin{aligned}{\frac {d^{2}\mathrm {x} _{1}}{dt^{2}}}&={\frac {k^{2}\mathrm {M} }{{\boldsymbol {r}}^{3}}}(1-\mu )\left\{-{\boldsymbol {x}}+\mu \lambda ^{2}{\boldsymbol {\frac {x}{r}}}+\left(\xi _{1}-\xi _{2}\right){\boldsymbol {r}}\epsilon _{1}+{\tfrac {3}{2}}{\boldsymbol {x}}\epsilon _{2}^{2}+{\boldsymbol {x}}\phi _{1}^{2}\right\},\\{\frac {d^{2}\mathrm {x} _{2}}{dt^{2}}}&={\frac {k^{2}\mathrm {M} }{{\boldsymbol {r}}^{3}}}\mu \left\{{\boldsymbol {x}}+(1-\mu )\lambda ^{2}{\boldsymbol {\frac {x}{r}}}+\left(\xi _{1}-\xi _{2}\right){\boldsymbol {r}}\epsilon _{2}-{\tfrac {3}{2}}{\boldsymbol {x}}\epsilon _{1}^{2}-{\boldsymbol {x}}\phi _{2}^{2}\right\}.\end{aligned}}$

(20)

In these equations the system of reference is arbitrary. We can always, by a Lorentz-transformation, make the velocity of any point equal to zero. If there were a point in the system having no acceleration, like the centre of gravity in Newtonian mechanics, the equations could be simplified by taking that point as origin of the system of reference. Now the integrals of the motion of the centre of gravity do not exist in this simple form under the new law, but it is still possible to find a point without acceleration. The centre of gravity has the coordinates $x_{0},\ y_{0}\ z_{0}$, defined by—

$x_{0}=\mu x_{1}+(1+\mu )x_{2},\dots$

We have thus

$\xi _{0}=\mu \xi _{1}+(1-\mu )\xi _{2}.$

If we denote relative velocities by letters without index, thus—

$\xi =\xi _{1}-\xi _{2},$

$\phi ^{2}=\xi ^{2}+\eta ^{2}+\zeta ^{2}$, ${{\boldsymbol {r}}\epsilon }={{\boldsymbol {x}}\xi }+{{\boldsymbol {y}}\eta }+{{\boldsymbol {z}}\zeta },$

etc.

We have

$\xi _{1}=\xi _{0}+(1-\mu )\xi ,$ $\epsilon _{1}=\epsilon _{0}+(1-\mu )\epsilon ,$

$\xi _{2}=\xi _{0}-\mu \xi ,$ etc.

We find at once from (20)—

${\frac {d^{2}x_{0}}{dt^{2}}}={\frac {k^{2}\mathrm {M} \mu (1-\mu )}{{\boldsymbol {r}}^{3}}}\left\{-(1-2\mu )\lambda ^{2}{\boldsymbol {\frac {x}{r}}}+{\boldsymbol {r}}\xi \left(\epsilon _{1}+\epsilon _{2}\right)-{\tfrac {3}{2}}{\boldsymbol {x}}\left(\epsilon _{1}^{2}-\epsilon _{2}^{2}\right)+{\boldsymbol {x}}\left(\phi _{1}^{2}-\phi _{2}^{2}\right)\right\}.$

(21)

The right-hand member is of the second order, and is thus the same for the laws I. and II. In it we can, for the relative coordinates and velocities, use their values in ordinary Keplerian motion. If then we take the orbital plane for the plane of ($x,y$) and the axis of $x$ through the perihelion, the equations (21) become—

${\begin{aligned}{\frac {dc\xi _{0}}{dt}}&=\mathrm {P} +\mathrm {Q} .c\xi _{0}+\mathrm {R} .c\eta _{0}\\{\frac {dc\eta _{0}}{dt}}&=\mathrm {P} '+\mathrm {Q} .c\xi _{0}+\mathrm {R} '.c\eta _{0}\\{\frac {dc\zeta _{0}}{dt}}&=0.\end{aligned}}$